help-39

[grave parcel]

Would the angle in this case be 55

?

For question 9

[cloud zephyr]

Recall: Angle in True bearing is the angle measured from the north, clockwise

[grave parcel]

Ok

I get that

[cloud zephyr]

Side Note: North East South West are all 1 right angle (90°) apart

[grave parcel]

Is this what u mean?

[grave parcel]

Yes I get that

[languid spindle]

boi wot da hell happen here

[grave parcel]

Excuse me?

[cloud zephyr]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[languid spindle]

mb

[cloud zephyr]

now that we have a graph

[grave parcel]

Ah yes I see

[cloud zephyr]

we wanna find how far east did ironman travel

[grave parcel]

Where does that 35 degrees come from

[cloud zephyr]

it's from 125°-90°

[grave parcel]

Oh to find the angle u do true bearing - 90?

So like if I have a true bearing of 310

I would do 310 - 90 to find the angle?

[cloud zephyr]

nah, it depends on how large it the angle (what quadrant it is in)

[grave parcel]

I don’t get that bit

Like if I can do that

[cloud zephyr]

now we can construct the green triangle, and using sin/cos/tan to find the distance due east

[grave parcel]

Then I can do compass bearing

[cloud zephyr]

true

[grave parcel]

Tan

[cloud zephyr]

note that compass bearing and true bearing are 2 different measurements, but they can tell the same angle with appropriate numbers and signs

[grave parcel]

What’s the difference?

[cloud zephyr]

lemme give some examples

it'll take some time

[grave parcel]

It’s fine

Let’s just continue with the question

[cloud zephyr]

oh. ok

[grave parcel]

Thank you tho

[cloud zephyr]

now that we have a green traingle

we check out which one is hypotenuse first

recall: for Right-angled triangle, the Hypotenuse is the line directly opposite to the right angle

[grave parcel]

I know trigonometry

We are using tan

Wait

[cloud zephyr]

and as i have shown (in red) the right angle, we have 3.75km is the hypotenuse

[grave parcel]

Nvm

I don’t see the hypotenuse

lol it’s sin

[cloud zephyr]

since we need to find "due east"

we should use cosine

[grave parcel]

Hmmmm

[grave parcel]

Isn’t east the opposite?

[grave parcel]

Oh my days

HOW DID I MESS UP SO BADLY

3.07 is the true bearing?

[cloud zephyr]

nope? what is 3.07

[compact ridge]

,w 3.75 * cos(35*pi/180)

[grave parcel]

3.07 degrees?

[compact ridge]

you're using the wrong trig ratio

[grave parcel]

Huh

[compact ridge]

Which side is east?
(adj/opp/hyp)

[grave parcel]

Adjacent

[compact ridge]

yes

And 3.75 is our hypotenuse

[grave parcel]

Yes

[compact ridge]

So SOHCAHTOA, we need cos

[cloud zephyr]

and 35° is our theta

[grave parcel]

So 3.75 x cos(35)

[compact ridge]

oh waitttt

[grave parcel]

Is 3.07

[cloud zephyr]

yea, let's check

[compact ridge]

yeah okay yes it's 35 ofc

[jolly parrot]

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Requested by sooouth.

[compact ridge]

sorry I misread the question and 3.07 m is correct

[grave parcel]

How about Scooby doo question

[compact ridge]

Remember we wanted 'how far east' so this has to be a distance

[compact ridge]

Ah okay, can you try drawing it out?

[grave parcel]

I’ll try one sec

[compact ridge]

It might help to draw a 45-90-45 triangle to figure out how many metres south 600m SW is

And how west 600m SW is also

[grave parcel]

Idk how to draw it

.close

[pearl pond]

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[pearl pond]

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[compact ridge]

It really depends on what u is

[pearl pond]

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[upper crypt]

This, sorry

[compact ridge]

ohhhhh

Okay so u is not constant

[upper crypt]

Yeah

[compact ridge]

What's the original problem?

[upper crypt]

This, but instead of u it said ln(x)

And instead of u tends to infinity, it was x

[compact ridge]

Hmmmm ok

[upper crypt]

Can you use L'hopital's?

[compact ridge]

You should really tidy this up first though, so

$$\frac{\sqrt{x}^{\sqrt{u}} \sqrt{x}^{\sqrt{u}} \sqrt{u}^x}{\sqrt{x}^u \sqrt{u}^{\sqrt x} \sqrt{u}^{\sqrt x}}$$

[jolly parrot]

south

[upper crypt]

True

[compact ridge]

Basically rewrite $x^{\sqrt u} = (\sqrt{x}^2)^{\sqrt u} = \sqrt{x}^{2 \sqrt u}$

[jolly parrot]

south

[compact ridge]

And then we get $\sqrt{x}^{2 \sqrt{u} - u} \sqrt{u}^{x - 2 \sqrt{x}}$

[jolly parrot]

south

[upper crypt]

Wait what

[compact ridge]

Or $x^{\sqrt{u} - u/2} u^{x/2 - \sqrt{x}}$

[jolly parrot]

south

[upper crypt]

???

Brb

[compact ridge]

laws of indices and stuff

[tulip mason]

Just replace x with u.

So for instance:
[x^{\sqrt{u}}=(e^{u})^{\sqrt{u}}=e^{\sqrt{u^3}}]

You should be able to write everything in terms of powers of e, you might need to use $e^{\log(x)}=x$.

[jolly parrot]

Max

[upper crypt]

Ah okay ty

[compact ridge]

And sqrt(u) - u/2 > 0, so if we replace x^(...) with x^0 = 1 this will be smaller
But x/2 - sqrt(x) > 0 for x large enough, and ln x is increasing, so this goes to +infinity overall

[pearl pond]

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[upper crypt]

Another limit question

How do I get the following limit in a form such that L'Hopital's rule is applicable?

$\lim_{x\to0^{+}}((\frac{sin(x)}{x})^{\frac{1}{x^2}})$

[jolly parrot]

Josh ♘

[upper crypt]

@tulip mason?

[tulip mason]

Use $e^{\log(x)}=x$

[jolly parrot]

Max

[upper crypt]

How does that help here?

[compact ridge]

Yes you will get a 0/0 form
Consider ln L = (1/x^2) ln(sin(x) / x) by logarithm rules

As in take the ln of the limit

[upper crypt]

Ah okay

[compact ridge]

Then sin(x) / x will go to 1, and ln 1 will go to 0

[upper crypt]

So if L is the limit

I can log both sides but put the log in the limit?

[compact ridge]

Yes so once you find ln L, you can do e^(ln L) to recover L

[upper crypt]

Okay ty

[compact ridge]

npnp

[pearl pond]

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[pearl pond]

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[balmy scaffold]

how do i solve this

for x

[old marsh]

$x^2 - \sqrt x = 0$

[old marsh]

whatve u tried @balmy scaffold

[balmy scaffold]

yeah but how to solve for x no

[old marsh]

yes we will get there

[balmy scaffold]

do you put it into the p q formula

[old marsh]

no

try factoring out the gcf

[balmy scaffold]

x(x - sqrt 1 )?

[old marsh]

not quite

if u distribute that, youll see u dont get the original expression

[balmy scaffold]

ye no idea how to factor out squaroots

[old marsh]

try factoring out sqrt(x)

@balmy scaffold aight brodie

[balmy scaffold]

oh sec

[balmy scaffold]

yeah no idea how to do that

[old marsh]

$x^{0.5}(x^{?} - 1) = 0$

[balmy scaffold]

ehm

[old marsh]

what goes in the question mark part

[balmy scaffold]

sec

also 0.5

since they have to get added to make 1

or wait

we had x^2

so it must be x^1.5

[old marsh]

yep

so now we have

$x^{0.5}(x^{1.5} - 1) = 0$

[old marsh]

now u can set each component equal to 0, and solve for x

[balmy scaffold]

x is the only component?

dunno how to solve this

1.5sqrt(1)?

[old marsh]

$x^{0.5} = 0 \newline \newline x^{1.5} - 1 = 0$

[old marsh]

solve those

[balmy scaffold]

oh i haven't seen this type of style being used before

[old marsh]

wdym?

[balmy scaffold]

x is 0 in the first one i guess

[old marsh]

yes but do u know why

[balmy scaffold]

in the second one it must be 1.5sqrt(1)

[old marsh]

1.5sqrt(1) is just 1.5.

if u put that in for x, does the equation work?

[balmy scaffold]

geez calculating that in itself is a chore

[old marsh]

,calc 1.5^(1.5) - 1

[jolly parrot]

Result:

0.83711730708738

[old marsh]

doesnt work

so heres the thing

for equations like this and all other equations not like this

we wanna get all terms with an x to one side, and all terms without an x to the other side

so for the second equation, we get

$x^{1.5} = 1$

[old marsh]

now from here

how can we get rid of that 1.5 exponent

what should we do to both sides

[balmy scaffold]

1. 5 sqrt()

[old marsh]

by 1.5 sqrt, are u referring to

[balmy scaffold]

on both sides we take the 1.5sqrt

[old marsh]

$\sqrt[1.5] x^{1.5} = \sqrt[1.5] 1$

[old marsh]

that?

[balmy scaffold]

zeah i thought that would solve it

[old marsh]

ahhh ok i misunderstood u at first

thats correct

thats what we do

so what does that equation simplify to

[balmy scaffold]

1.5?

[old marsh]

x = 1.5 ?

[balmy scaffold]

ye that's why you said earlier

[old marsh]

we tried this here and it didnt work

[old marsh]

1.5sqrt(1) means 1.5 * sqrt(1)

it doesnt mean 1.5 is in the index

[balmy scaffold]

i thought you meant it that way

but yeah how to you calculate this

[old marsh]

no, we dont represent it like that

we can rewrite that as

$1^{1.5} = 1^{\frac 32}$

[balmy scaffold]

ah ok intersting

[old marsh]

so now we can do a few things

[balmy scaffold]

so we have 0 as 1 solution

and as the second

[old marsh]

$1^{\frac 32} = (1^3)^{0.5} = (1^{0.5})^3$

[old marsh]

if u simplify whats in the parentheses and then apply the outside exponent, what do u get

[balmy scaffold]

eh

that would be another complicated one with sqrt(1)^3

[old marsh]

bro

no

[balmy scaffold]

how else to simplify it

[old marsh]

which one do u wanna go with, the 2nd or third expression?

[balmy scaffold]

i guess the second is the correct one

[old marsh]

both are correct

its just a matter of which order u wanna go in

so for the second one

whats 1 cubed

[balmy scaffold]

1

[old marsh]

ok so now we have

$1^{0.5}$

[old marsh]

whats another way of writing that

[balmy scaffold]

sqrt(1)

[old marsh]

and whats that equal to

[balmy scaffold]

1

[old marsh]

yea

[balmy scaffold]

ah interesting

[old marsh]

bro u really need to review ur algebra

what about the

$x^{0.5} = 0$

[balmy scaffold]

yeah i forgot how to deal with log and sqrt

[old marsh]

what would u do to both sides

[balmy scaffold]

sqrt

[old marsh]

no, u dont sqrt both sides

[balmy scaffold]

and that is why it is 0 i guess

what

[old marsh]

its already a square root

why would u square root again

$\sqrt{x} = 0$

[balmy scaffold]

ohhh

[balmy scaffold]

^0.5

[old marsh]

no

^0.5 is just square root

[balmy scaffold]

on this one to get x i would just ^0.5 no?

[old marsh]

no

if u ^0.5 both sides, u would get

$x^{0.25} = 0^{0.5}$

[balmy scaffold]

oh so i have to ^2

[old marsh]

yes

heres the thing

whenever u have an equation of the form

$x^{a} = c$

[old marsh]

to solve for x

raise both sides to the power of 1/a

$(x^a)^{\frac 1a} = c^{\frac 1a}$

[old marsh]

and that cancels out and becomes

[balmy scaffold]

ah interesting

[old marsh]

$x = c^{\frac 1a}$

[balmy scaffold]

yeah i see it

1/0.5 = 2

[old marsh]

yes exactly

[balmy scaffold]

ah so since we have 2 components we have 2 intersections

one at 0 and other one at 1

[old marsh]

yea

,w x^2 - x^(0.5) = 0

[jolly parrot]

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Requested by northsteve

[old marsh]

theres the graph

[balmy scaffold]

interesting

i'll try to get the area between the 2 functions now thx for the help

[pearl pond]

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[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

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[dense plover]

For the first part is this correct?

[rancid rune]

lol they literally give u the answer

you dont need to subtract the vectors

[dense plover]

Oh💀

[rancid rune]

you already have the coplanar vectors

if a and b were points on the plane, then you would subtract their position vectors to find the coplanar vectors

[dense plover]

Oh ok

Ty

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[dusky pagoda]

in tournament systems, how do you calculate the bye?

[dusky pagoda]

my teacher said "the formula of bye is the power of 2 minus the number of teams" and that kind of confused me

is it 7^2 or do i keep multiplying 2 to itself until i find a number higher than the number of teams?

.close

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[jovial grove]

Hi. I’m trying to show that this inequality leads to epsilon with the prescribed inequality on k, but I am ending up with 2*epsilon instead.

[jovial grove]

I went ahead and replaced z_k with N_f^k(z).

Oh. I am also assuming base 2 on the logarithms.

[dire parrot]

nvm forget what i said

[pearl pond]

@jovial grove Has your question been resolved?

[dire parrot]

since the issue here is with epsilon <= 1, then youd have to assume some k' > log(|log(2^p * epsilon)| + log(M)) such that 2^p * epsilon >= 1 (which is guaranteed), so then itll work in both cases for some random epsilon > 0

[pearl pond]

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@smoky spoke Has your question been resolved?

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[fresh silo]

hello

[fresh silo]

is this true or no ?

[pearl pond]

@fresh silo Has your question been resolved?

[fresh silo]

<@&286206848099549185>

[dry thorn]

is this binomial thingy

[fresh silo]

yes

[pearl pond]

@fresh silo Has your question been resolved?

[pearl pond]

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[hoary meadow]

i dont understand how the existence of a potential energy must imply total energy is conserved

[sharp vigil]

the change in kinetic energy is given by the work done by a force (by work energy theorem). if the force is conservative, then all work done by it can be found in terms of the potential energy, so the sum of potential and kinetic energy must be constant

[hoary meadow]

im still kinda confused what exactly is potential energy

and if all work done can be found in terms of the potential energy then how does that imply the sum of the potential energy and ke is constant

[sharp vigil]

we have by work energy theorem that for kinetic energy $K$ the work done ($W$) is related by [ \Delta K = W ] If a potential energy function exists, then we have that $W = -\Delta U$. then we have that [ \Delta K = - \Delta U \implies \Delta K + \Delta U = \Delta(K + U) = 0 ] in other words, the sum $K + U$ overall does not change.

[jolly parrot]

cloud

[hoary meadow]

ohhh

wait but

cant like

a particle still have potential energy

but it's not the case energy is conserved

?

[sharp vigil]

well a potential energy is only associated with a particular force. any non-conservative forces would do work that isn't described by a potential energy function (so energy wouldn't be conserved), but any changes to the energy of the particle must be caused by work due to some force

[hoary meadow]

wait why is work done = -change in potential energy

@sharp vigil

[sharp vigil]

that's how we define potential energy

the - sign is precisely so that the sum of kinetic and potential energy is constant as shown

[hoary meadow]

wait then why does it follow that work done is also = change in ke

if we define work done as -change in potential energy

[sharp vigil]

the work-energy theorem says that the work done = change in KE for any force (conservative or not)

in general the work done is the line integral $W = \int_a^b \vb F \cdot \odif{\vb r}$, and this is always equal to the change in kinetic energy

[jolly parrot]

cloud

[sharp vigil]

however if the force is conservative, we have by the fundamental theorem of line integrals that if there exists some potential function $f$ such that $\vb F = \nabla f$, then the line integral [ W = \int \vb F \cdot \odif{\vb r} = \Delta f ] we define the potential energy to be the negative of this function: $U = -f$

[jolly parrot]

cloud

[pearl pond]

@hoary meadow Has your question been resolved?

[pearl pond]

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[covert pivot]

So all you want is the height?

[steel canopy]

trigonometry?

[covert pivot]

14sin(46)

[pearl pond]

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[smoky saffron]

can someone please explain what is going on

this is all i got so far 😭

[hasty osprey]

try resolving both vectors A and B into rectangular components (x and y components)

[smoky saffron]

Okay

Will you stay here as i do it

[hasty osprey]

sure

[smoky saffron]

before i continue to do a's component is this correct?

[hasty osprey]

yea looks good

[smoky saffron]

what now?

do i calculuate the magnitude of B's y and x ?

[hasty osprey]

vector B points left and vector A points right

[smoky saffron]

what do you mean

like their arrows

[hasty osprey]

yeah

[hasty osprey]

so u have components of a

and components of b

find the resultant vector by adding the components

[hasty osprey]

pointing rightwards up

not rightwards down

[smoky saffron]

wait what unit are these

[hasty osprey]

doesnt rly matter because ur trying to find angle

[smoky saffron]

oh

its the magnitude tho right

[hasty osprey]

as its force, likely newtons but doesnt matter

[hasty osprey]

ur components for B are Fx = 2.819 (left) and Fy = 1.026 (up)

ur components for A are Fx = 0.866 (right) and Fy = 0.5 (up)

[smoky saffron]

yes

[hasty osprey]

ur resultant in the y direction is 1.026 + 0.5

[smoky saffron]

oh?

[hasty osprey]

ur resultant in the x direction is 2.819 - 0.866

tell me clearly what u dont understand and ill explain

[smoky saffron]

what do you mean resultant in the y direction

of what

[hasty osprey]

the components

u found 2 y components, i can find the resultant y component by adding them as theyre both up

[smoky saffron]

ohhh

ohhhh

ok

how come you add the y's but subtract the x's

[hasty osprey]

^

the y components are both up

but the x components, one is left and one is right

so we have to subtract this time to find the resultant

[hasty osprey]

??

[smoky saffron]

I understand this but i dont understand how you kne w they were going left or right

[hasty osprey]

the question tells u

[smoky saffron]

oh

okay

[hasty osprey]

^

[smoky saffron]

anyways what do we do now that we found the resultants

[hasty osprey]

what was ur x and y resultants

and state direction

[smoky saffron]

y resultant is 1.526 up

[hasty osprey]

yup

[smoky saffron]

x resultant is 1.953

uh

idk which way

right?

bc its 3x stronger

[hasty osprey]

we had 2.819 which was left and 0.866 right..

[smoky saffron]

oh so left

[hasty osprey]

we subtracted 0.866 from 2.819 so its left

[hasty osprey]

left is 3x stronger

read the question carefully

[smoky saffron]

okay your right

have a question tho

[hasty osprey]

alr cool

oh yea ask

[smoky saffron]

what are these diagrams that this person did?

[hasty osprey]

creating triangles to likely use sine rule/cosine rule

lot more complicated than it needs to be

[smoky saffron]

wait im confused tho cuz we didnt find an angle

[hasty osprey]

we’re not done

[smoky saffron]

oh

[hasty osprey]

all we did is subtract and add some numbers so far

so we have 1.953 left and 1.526 up

draw those two vectors

[hasty osprey]

now find the angle the triangle makes with the vertical

[smoky saffron]

Im confused how do i do it

[hasty osprey]

just trig?

do u know trig ratios?

[smoky saffron]

not rly buut if u explain i might remember

[hasty osprey]

sin, cos, tan

sin is opposite/hypotenuse

[smoky saffron]

ohh soh cah toa

[hasty osprey]

cos is adjacent/hypotenuse

yeah

trig ratios

[smoky saffron]

ohh

[hasty osprey]

so if u call the angle theta

[smoky saffron]

im finding this angle?

[hasty osprey]

and use inverse trig ratios u can solve

[hasty osprey]

well, it should be the angle it makes with the boat

[hasty osprey]

yeah so from the boat

u can find the angle u marked first, then ill show what to do

[smoky saffron]

I did tan inverse and got 37.94?

they got 38.4

whyd they do 38.4 - 20 tho

[hasty osprey]

its due to small rounding, 37.94 and 38.4

[smoky saffron]

yes yes

but what about the 38.4 - 20 why did they do that

[hasty osprey]

u made a small mistake

@smoky saffron

[smoky saffron]

I just realized

my whole diagram was wrong

[hasty osprey]

yeah

angles

[smoky saffron]

Im redoing it

[hasty osprey]

fix up the angles and show me before u do any resultant calcs

[hasty osprey]

redraw

draw A and B in the correct direction too

[smoky saffron]

no this is correct

cuz look

[hasty osprey]

its not correct

[hasty osprey]

this diagram is right

but not ur triangles

[smoky saffron]

oh

like my components

[hasty osprey]

for B, the angle should be 60 degrees

and A should be 70 degrees

inside the triangle

[hasty osprey]

that 30 should be a 60

cus look at the diagram

it makes a 30 degree angle with the vertical line in the middle

[hasty osprey]

this triangle

the 30 should be a 60

[smoky saffron]

ohok

[pearl pond]

@smoky saffron Has your question been resolved?

[pearl pond]

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[wind meadow]

what has gone awry

[lean saddle]

since both x^2 and 1+x^2 are irreducable quadratics (over R) you would need ax+b in the numerator of each

(diff leters for both)

[wind meadow]

waiiit

since they're squared?

or what does irreducible mean

[plush oriole]

That it cannot be reduced.

[lean saddle]

cant be factored any more

like if you had x^2 - 3x + 2 you could factor it more

to get (x-2)(x-1)

but with x^2 + 1 you cant really do that with out complex numbers

[wind meadow]

so since it's squared I put the x on the top?

[lean saddle]

kind of yes

[wind meadow]

and if it's x cubed do i put x^2 on the top?

im confused how faactoring has to do with x

[lean saddle]

if its cubed then it will always have at least one real root

[lean saddle]

do you want to learn?

[wind meadow]

yes

[lean saddle]

ok so lets suppose you have just one term

suppose 1/(x-1) and you wanted to factor it

[wind meadow]

yeaa

[lean saddle]

you would obviously just do 1/(x-1) = a/(x-1) and then just say a = 1

right?

cuz theres really no other choice

[wind meadow]

yeah

[lean saddle]

now lets say we had two terms

and we knew the roots already

so we had 1/((x-a)(x-b))

[wind meadow]

yeeah

[lean saddle]

then we would ideally want it as A/(x-a) + B/(x-b) right?

[wind meadow]

right

[lean saddle]

real quick do you know what complex numbers are?

[wind meadow]

numbers that dont exist right?

[lean saddle]

uh

well

[pseudo grail]

lol

[lean saddle]

no ish

$ i := \sqrt{-1} $

[dapper kraken]

a+bi

[lean saddle]

have you see this i before?

[dapper kraken]

$i := \sqrt{-1}$

[jolly parrot]

Skill_Issue

[wind meadow]

imaginary?

yeah

[lean saddle]

ok this makes it much easier

there only one rule you need then

[balmy helm]

what was the answer to that rectangle paper folded olympiad problem

[dapper kraken]

WHAT THE FUCK uh idk

[lean saddle]

so have you learnt how every polinomial has n roots?

[wind meadow]

XD

[lean saddle]

^

[wind meadow]

i think so

this is related to how x^2 - 4 equal both x + 2 and x - 2 right?

[lean saddle]

yes

so if we have somthing like x^2 -1 = 0

then we would have?

[wind meadow]

x = 1, -1

[lean saddle]

yup

and if it were x^2 +1 =0?

[wind meadow]

x = i?

[lean saddle]

there is one more

can you try to find it?

[wind meadow]

x = -i

[lean saddle]

yup

so we have (x-i)(x+i) = x^2 +1 right?

[wind meadow]

right yeah

[lean saddle]

so lets go back to 1/(x^2+x^4)

try to factor it as much as possible

(consider 0 a factor for now)

[wind meadow]

I can only see x^2 (1 + x^2)

[lean saddle]

but we say that x^2 + 1 can be factored a little bit more right?

[wind meadow]

ohh right (x+i)(x-i)

[lean saddle]

we can also write x^2 as (x-0)(x+0)

[wind meadow]

ooh i see

[lean saddle]

so what are the 4 factors?

[wind meadow]

0, -i, i anddd -0?

[lean saddle]

yup

so we have 1/(x^2+x^4) = A/(x-0) + B/(x+0) + C/(x-i) + D/(x+i)

just from making them all linear factors

[wind meadow]

Oooooh i understand now

[lean saddle]

but the complex numbers and the 0 kinda make it look a little wierd right?

[wind meadow]

right yea

[lean saddle]

so what if you try to mix A/(x-0) + B/(x+0)

try to combine them into one term

[wind meadow]

Ax + Bx

oh wait thats not divided

wait so we're turning the A(x-0) + B/(x+0) into A/(x-0)+B(x+0)?

[lean saddle]

i was kinda hoping we could reach somthing like (a+bx)/(x-0)^2

[wind meadow]

wait so how come A isnt a fraction here?

dont we normally turn it into a fraction?

[lean saddle]

oh it is meant to be

i just missed the /

[wind meadow]

oh and C too?

[lean saddle]

yup

i fixed it now

[wind meadow]

oooh okay

[lean saddle]

sorry about that

[wind meadow]

no problem

[wind meadow]

is B supposed to be a fraction here?

[lean saddle]

yes

my bad

[wind meadow]

ooh okay np

so then it'd be Ax + Bx/(x-0)(x+0)?

[lean saddle]

yeah in this case pretty much exactly that

now lets try somthing simmilar with C/(x-i) + D/(x+i)

[wind meadow]

oo itd be Cx + Ci + Dx - Di/(x-i)(x+i) right?

[lean saddle]

im just going to write that but factored a little bit

(Cx + Ci + Dx - Di)/(x-i)(x+i)
(Cx + Ci + Dx - Di)/(x^2+1) , from what we used to get these
((C-D)i + (C+D)x)/(x^2+1)

C-D is just some other number right?

and C+D is also just some other number

[wind meadow]

yeeah

[lean saddle]

so lets call C-D = F

[wind meadow]

Sure yah

[lean saddle]

and C+D = G

(Fi +Gx)/(x^2+1)

but then Fi is also just some number

so lets call Fi = H

(H+Gx)/(x^2+1)

do you see how we get these x terms appearing naturally?

so putting it all together we have essentially, ill use completly new letters

[wind meadow]

yeah i see

[lean saddle]

J/x + K/x^2 + (L+Nx)/(1+x^2)

[wind meadow]

wait so where did L and N come from?

[lean saddle]

here

i just wanted completly new letters

[wind meadow]

ahh okay

[lean saddle]

does that make sense?

[wind meadow]

Yeah, but how does this tie into the Ax+B from the start?

[lean saddle]

which part?

reply to the message

[wind meadow]

this is the same as

well nobody posted it yet but

the solved problem has only Ax + B right?

so how does J, K, L and N tie into those?

[lean saddle]

a and b are just letters

so are j k l n

[wind meadow]

so if i were to actually solve this would i end up with complex numbers?

[lean saddle]

nope

[midnight haven]

Partial fractions?

[lean saddle]

yeah

just trying to show the derivation

[lean saddle]

try solving it

youll see the numbers and itll be good practice

[wind meadow]

okay will do rn

so to make sure i dont make any mistakes, the form would be Ax/x^2 + B/1+x^2?

[lean saddle]

itll be of thise form remember?

[wind meadow]

ooh okay i see

I end up with Jx^2+Jx^4+Kx+Kx^3+Lx^3+Nx^4 = 1

[pearl pond]

@wind meadow Has your question been resolved?

[pearl pond]

@wind meadow Has your question been resolved?

[pearl pond]

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[heady finch]

Can anyone explain how to get to the result in question 2?

I'm confused about the step of isolating x to gain x^2 in this case

since expanding you obtain (x^2-6x+9)+y^2=1

The reason I'm attempting to isolate is to be able to put the form into a format that can be solved with disc integration (integral of pi(x^2)dy since this is rotation in y axis)

[pearl pond]

@heady finch Has your question been resolved?

[midnight haven]

Do you know about volume of revolution formula?

[heady finch]

integral from bounds of a to b pix^2dy or piy^2dx depending on which axis

I was assuming because this is revolution about y so we needed to find pix^2dy where x is f(y)?

[midnight haven]

Maybe trig substitution?

[heady finch]

mmmm

I haven't learnt trig substitution yet

my apologies

[midnight haven]

Ah alright

It should be simpler then

[heady finch]

it's not the usual textbook I use so it might be in there

the current textbook I use doesn't even go over arcsin which is involved in the final integral's solution

if my substitution into wolframalpha was correct

[midnight haven]

Yeah well this integral have different bounds 1 and 0

So if initially we have 0 to 2pi

[heady finch]

the integral is meant to be derived from the circle though, unless I'm mistaken

[midnight haven]

Beuh

I was

Using wrong

Equation

We want x^2

https://youtu.be/DtVpk1b92IM?feature=shared

[heady finch]

yeah that's what I assumed but I was wondering how to isolate for the expansion

[midnight haven]

I don't get what's happening with boundaries

[heady finch]

if you expand to (x^2-6x+9) I was wondering how to get rid of the x on one side in this case

[midnight haven]

You can get rid of x

But

Boundaries

Wont

Change

X=sqrt1-y^2.+3

But boundaries

Different

And we will have 3 intwgrals

Ah

Yeah

Well we can try to complete the square

But boumdaries

Bruh brhuh.

Can you give me any reference for this textbook?

[heady finch]

mmmm

Cambridge mathematics yr 11 unit 3 apparently

[midnight haven]

Wait

0 to 1

Is vertical

Distance

Radius

So that's would make sense if they summing up areas in different way

[midnight haven]

Thanks

[midnight haven]

This?

@heady finch

Which page?

Found it

Alright

Give me some time I will try to figure it out

[heady finch]

tyty

[heady finch]

that's the one

pg 439 or around

[midnight haven]

Ah

[heady finch]

figured something out?

[pearl pond]

@heady finch Has your question been resolved?

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[midnight haven]

No sorry, I will try to do it later, ( I am busy). I will dm you when I finish.

I don't see terms cancelling out.
I tried to use the fact that area of any circle with radius 1 will be the same regardless of the coordinates but it haven't really worked

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[stark void]

In a parallelogram, the acute angle is 45 . Find the area of a parallelogram
if its diagonals are 6 cm and 8 cm.

can someone help me with this tasks i quite don't understand it

[thin sigil]

Try to draw a sketch of the sitution

and remember the properties of paralelogram angles

[stark void]

i've got this

what do i do next

how do i find the angle between diagonals knowing knowing only angles of a parallelogram?

[thin sigil]

There's another property for angles and diagonals

for paralelograms

[stark void]

what property?

[thin sigil]

or rather for lengths of diagonals

[stark void]

so what do i do?

[thin sigil]

oh you got that already ok

[stark void]

wdym

[thin sigil]

don't you think you drew that sketch wrong though?

Like, the angle of 45 and 135 should be reversed

[stark void]

why would that matter

that's just the sketch

the important stuff is the values

[stark void]

about the point of intersection dividing the diagonals by 2?

[thin sigil]

ye

You need to find the surface area right? How could you express the surface area?*

[stark void]

do diagonales divide the angles by 2?

[thin sigil]

not neccesarily

There's an easy way to get the angle splits, you just have to see it properly

[stark void]

can you just tell me pls?(

[thin sigil]

Look at the paralelogram like this

focus on one diagonal

the bottom angle is 45

the right one is x

knowing that the full top angle is 135

what's would the part of the top angle be then?

I worded this weirdly lol

Oh actually this doesn't work out nice, hold on

[stark void]

i was told that the answer is 7

how do we get 7

by using the formula S = 1/2d1d2sina, where a is the angle between diagonales

we get that

7 = 24sina

sina = 7/24

[thin sigil]

Oh

you know the law of sines right?

[stark void]

yes

[thin sigil]

Try to apply the law of sines to this triangle to find a relationship between x and y

since we only need the sin of that angle in the middle maybe you can find a way to get it like that

[stark void]

can we write x in terms of y?

it will be

i don't think so

argh

[thin sigil]

Try now writing law of sinces

sines

for the other triangle as well

that the diagonals form

This one

[stark void]

oh

i see

[thin sigil]

You should get something like this

solve this system to get x and y

Solving this analitically seems kinda tough, but probably managable

But you might be able to avoid it altogether, this the angle in the middle of diagonals can be written as x+y, it's sine being sin(x+y)

is what you're really after here

[pearl pond]

@stark void Has your question been resolved?

[thin sigil]

Solving it numerically I got this

[stark void]

yes

that's right answer

[thin sigil]

So basically solve this system, express sinx in term of siny from the first one

[stark void]

kind of too complicated

how should i have gotten it meh

[thin sigil]

it is, but it's doable

you'll probably have to solve a quadratic at some point

but yeah

it boils down to applying law of sines twice and then solving the system

maybe there's another approach, but I don't see it honestly

[stark void]

can't we substitute sinx/siny for 4/3?

and it'll be

sin(3pi/4 - x)/sin(pi/4-y) = sin(x)/sin(y)

nah that's too much

but 'kay

ty for yuor help legend!

really appreciate it

[thin sigil]

That would get you nowhere

[stark void]

yeah i've got it

[thin sigil]

you'd have to express sin(y) = 4/3 sin(x)

and then cosy with root

[stark void]

ik

[thin sigil]

and then plug that in, deal with the mess

and solve as a quadratic probably

[stark void]

yeah

it isn't very compicated

just

too long

but ty

[thin sigil]

yeah a lot of ground work

[stark void]

okay ty

.close

[pearl pond]

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[sand robin]

how can i show that set $M=\left{(x, y) \in \mathbb{R}^2 \mid x^2+4y^2=4\right}$ is bounded?

[jolly parrot]

Slowaq

[sand robin]

i know that it is and elipse

[thin sigil]

Do you just have to prove it's bounded or find it's bounds as well?

[sand robin]

just prove that it is bounded

[thin sigil]

I mean proof of contradiction works nicely here... clearly 9999^2 + 4 * 9999^2 isn't 4

assume it isn't bounded

find counterexample

hur dur

[sand robin]

ah alright i see it pretty easy proof

[thin sigil]

Probably nowhere nearly as rigorous as you could get

with my example

there's likely a much nicer way to go about it

I love plugging in stupid numbers in my functions to see what they do

[fluid axle]

how does that show anything really

[thin sigil]

That's just an example

you could prove that the set doesn't go through x = 99999

x = -99999

y = 9999 and y = -9999 each seperately

for example

and therefore it's bounded somewhere in that region

[fluid axle]

right so essentially the "ellipse" is inside the rectangle [-2, 2] x [-1, 1]

[thin sigil]

yeah but I like 9999

[fluid axle]

the rectangle is bounded so the ellipse is bounded

[sand robin]

cant i just say that set M is elipse with centre at origin and axis 4 and 2

[sand robin]

yes

[thin sigil]

yeah but you don't know that going into the problem

kinda

assume you don't know it's an elipse

Like, in the general case you don't know the shape, so best not to assume you do

This is for proving that it's bounded, now finding the bounds

that's a different story

There you can have some real fun

[pearl pond]

@sand robin Has your question been resolved?

[pearl pond]

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[flint shuttle]

if you show f(ab) implies ab=ba, does it work the other way around with G abelian means f is a homo

[flint shuttle]

like the iff is automatic

[tropic saddle]

not exactly sure what you mean. you still have to show both directions

[flint shuttle]

This is what I have. Is the opposite direction true if I just go from bottom to top

,rotate

[tropic saddle]

it could be good to elaborate what you do from the third to the fourth line. but yes, afterwards you can just read the proof from bottom to top and then thats a proof for the other direction

[flint shuttle]

oh ok thanks

.close

[pearl pond]

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[pearl pond]

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[abstract mauve]

can anyone help me what are it is looking for because I have done everything else right, but I don't know where they are getting the area from

[abstract mauve]

its just the 24

that I dont get

the rest I get

[brave schooner]

you have the area under the red line but you need the area over the curve and under the red line

[abstract mauve]

fuck yeah

.close

[pearl pond]

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[pearl pond]

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[sand robin]

does this system of equations have a solution?
[-\sin x\sin y=0]
[\cos x\cos y=0]

[jolly parrot]

Slowaq

[sand robin]

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

$\lim_{x\rightarrow\left(\frac{\pi}{4}\right)}\left(\frac{\left(4\sqrt{2}-\left(\sin\left(x\right)+\cos\left(x\right)\right)^5\right)}{1-\sin\left(2x\right)}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$\lim_{x\rightarrow\left(\frac{\pi}{4}\right)}\left(\frac{\left(4\sqrt{2}-\left(\sin\left(x\right)+\cos\left(x\right)\right)^5\right)}{\left(\sin\left(x\right)-\cos\left(x\right)\right)}\right)$

$\lim_{x\rightarrow\left(\frac{\pi}{4}\right)}\left(\frac{\left(4\sqrt{2}-\left(\sin\left(x\right)+\cos\left(x\right)\right)^5\right)}{\left(\sin\left(x\right)-\cos\left(x\right)\right)^2}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

now

[rancid rune]

prob best to combine the trig functions into one

[sharp smelt]

yeah

was thinking of doing that

$\lim_{x\rightarrow\left(\frac{\pi}{4}\right)}\left(\frac{\left(4\sqrt{2}-4\sqrt{2\ }\left(\sin\left(x+\frac{\pi}{4}\right)\right)^5\right)}{2\left(\sin\left(x-\frac{\pi}{4}\right)\right)^2}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rancid rune]

oh nvm

[sharp smelt]

the problem is the power 5

$\lim_{x\rightarrow\left(\frac{\pi}{4}\right)}\left(\frac{\left(4\sqrt{2}\left(1-\left(\sin\left(x+\frac{\pi}{4}\right)\right)^5\right)\right)}{2\left(\sin\left(x-\frac{\pi}{4}\right)\right)^2}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rancid rune]

write u = x-pi/4

makes things a lot easier

then the sin in the numerator becomes sin(u+pi/2) = cos u

then some simplification gets u the answer

[sharp smelt]

ok

let me try that

$\lim_{u\rightarrow0}\left(\frac{\left(2\sqrt{2}\left(1-\left(\sin\left(u+\frac{\pi}{2}\right)\right)^5\right)\right)}{\left(\sin\left(u\right)\right)^2}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

yeah

oh

[eternal tulip]

Idk if that helps

[rancid rune]

wait this isnt the same as

oh so many brackets

ok yes

now write sin(u+pi/2) as cosu

[sharp smelt]

ok

[rancid rune]

ignoring the factor: $\lim_{u\to 0}\frac{1-\cos^5 u}{\sin ^2 u}$

[jolly parrot]

Obotron

[sharp smelt]

$\lim_{u\rightarrow0}\left(\frac{\left(2\sqrt{2}\left(1-\left(\cos\left(u\right)\right)^5\right)\right)}{\left(\sin\left(u\right)\right)^2}\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

ah

[rancid rune]

im gonna go from this$\lim_{u\to 0}\frac{1-\cos^5 u}{\sin ^2 u}$

[jolly parrot]

Obotron

[sharp smelt]

now I could l'hopital this

[rancid rune]

no need

[sharp smelt]

or is there any other way

ok let me think

[rancid rune]

write sin^2 as 1 - cos^2

$\lim_{u\to 0}\frac{1-\cos^5 u}{1-\cos ^2 u}$

[jolly parrot]

Obotron

[sharp smelt]

thought of that, does that actually help though

[rancid rune]

yes because we can factorise

$\lim_{u\to 0}\frac{(1-\cos u)(1+\cos u + \cos^2 u +\cos^3 u +\cos^4 u)}{(1-\cos u)(1+\cos u)}$

[jolly parrot]

Obotron

[sharp smelt]

that works lol

but really nasty

thanks

[rancid rune]

nah

that factorisation is very common

[sharp smelt]

I've never factorised a 5th degree polynomial in my life IMO

but ok

[rancid rune]

not just 5th degree

any polynomial of th eform: $1-x^n$

[jolly parrot]

Obotron

[sharp smelt]

ah

ok

Didn't know that identity

thanks

..clsoe

[rancid rune]

i bet u did

[sharp smelt]

.close

[pearl pond]

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